Skip to main content
Search type Search syntax
Tags [tag]
Exact "words here"
Author user:1234
user:me (yours)
Score score:3 (3+)
score:0 (none)
Answers answers:3 (3+)
answers:0 (none)
isaccepted:yes
hasaccepted:no
inquestion:1234
Views views:250
Code code:"if (foo != bar)"
Sections title:apples
body:"apples oranges"
URL url:"*.example.com"
Saves in:saves
Status closed:yes
duplicate:no
migrated:no
wiki:no
Types is:question
is:answer
Exclude -[tag]
-apples
For more details on advanced search visit our help page
Results tagged with
Search options not deleted user 95670

A quantum observable is a measurable operator whose corresponding property of the state can be determined by some sequence of physical operations ("observation"), such as submitting the system to various electromagnetic fields and eventually reading a value. In systems governed by classical mechanics, any experimentally observable value can be shown to be given by a real-valued function on the set of all possible system states.

2 votes
1 answer
173 views

Is the generalized uncertainty principle dependent on the state of the particle?

The generalized uncertainty principle can be written as (where A and B are observables): $$ \sigma_A\sigma_B \geq \left| \frac{1}{2i}\langle [A,B]\rangle_\Psi \right| $$ But the average value of the commutator … Hence is it right to say that for some states the observables A and B will have to respect the uncertainty relation and in some other cases it won't? …
E. Morell's user avatar